Question

Question: The diameter D of a tube does affect the node at the open end of a tube. The end correction can be roughly approximated as adding \(D/3\) to \(l\) to give as an effective length for the tube in calculations. For a closed tube of length 0.55 m and diameter 3.0 cm, what are the frequencies of the first four harmonics, taking the end correction into consideration?

Short answer

The uncorrected frequencies are 156 Hz, 468 Hz, 780 Hz, 1092 Hz and the corrected frequencies are 153 Hz, 459 Hz, 765 Hz, 1071 Hz.

Step-by-step solution

Determination of frequencies of four harmonics

The uncorrected frequency can be calculated using the speed and the length of the tube. For the correct frequency, the effective length is more than one-third time of the diameter.

Given information

Given data:

The length of the tube is \(l = 0.55\;{\rm{m}}\).

The diameter of the tube is \(D = 3\;{\rm{cm}}\).

Find the uncorrected frequencies

The uncorrected frequencies can be calculated as:

Here, \(v\)is the speed of sound at room temperature.

For \(n = 1,2,3,4.\)

\(\begin{array}{l}{f_1} = 156\;{\rm{Hz}}\\{f_2} = 468\;{\rm{Hz}}\\{f_3} = 780\;{\rm{Hz}}\\{f_4} = 1092\;{\rm{Hz}}\end{array}\)

Thus, the uncorrected frequencies are 156 Hz, 468 Hz, 780 Hz, and 1092 Hz.

Find the effective length and corrected frequencies

Find the effective length of the tube.

\(\begin{array}{c}{l_{{\rm{eff}}}} = l + \frac{D}{3}\\ = 0.55 + \frac{{\left( {3\;{\rm{cm}}} \right)\left( {\frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)}}{3}\\ = 0.56\;{\rm{m}}\end{array}\)

The corrected frequencies can be calculated as:

For \(n = 1,2,3,4\).

\(\begin{array}{l}{f_1} = 153\;{\rm{Hz}}\\{f_2} = 459\;{\rm{Hz}}\\{f_3} = 765\;{\rm{Hz}}\\{f_4} = 1071\;{\rm{Hz}}\end{array}\)

Thus, the corrected frequencies are 153 Hz, 459 Hz, 765 Hz, 1071 Hz.